How do you differentiate #y=3cot(ntheta)#?

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Answer 1 · 2017-09-27T07:54:13

#(dy)/(d theta)=-3ncsc^2 ntheta#

we will need the chain rule

#(dy)/(d theta)=(dy)/(du)(du)/(d theta)#

#y=3cot(ntheta)#

#u=ntheta=>(du)/(d theta)=n#

#:.y=3cotu=>(dy)/(du)=-3csc^2u#

#:.(dy)/(d theta)=(dy)/(du)(du)/(d theta)#

gives

#(dy)/(d theta)=(-3csc^2u)xxn#

simplifying and substituting back

#(dy)/(d theta)=-3ncsc^2 ntheta#